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So our teacher claimed that if we have a Potential of the form $V(x)= x^\nu$ then the Energy is of the form $E={2\nu \over \nu+2}$ Can anyone break up the math for this problem?

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    $\begingroup$ Do you mean the ground state energy? If $\nu$ is an odd integer the energy is not bounded below. $\endgroup$ – Keith McClary Oct 8 '15 at 3:32
  • $\begingroup$ Yes the bound state energy $\endgroup$ – Siddhartha Dam Oct 8 '15 at 4:31
  • $\begingroup$ $x^\nu$ in imaginary if $x<0$ and $\nu$ is $\frac{1}{2}$. $\endgroup$ – Keith McClary Oct 8 '15 at 4:42
  • $\begingroup$ V actually varies with x in,this manner.. So you can ignore the non trivial solutions $\endgroup$ – Siddhartha Dam Oct 8 '15 at 9:55
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Assuming you consider only even powers of $x$ (for odd powers there are no bound states) the behaviour you describe comes from the Born-Sommerfeld condition. This is described in the Wikipedia article Old quantum theory, and there is a related question on this site.

Suppose $E$ is the energy of your system, and $\pm X$ are the turning points where $E = V(X)$, then for your system the Born-Sommerfeld condition states:

$$ \int_{-X}^{+X} p \, dx = nh $$

where the momentum $p$ is given by:

$$ p(x) = \sqrt{2m(E - V(x))} $$

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